Problem: Let $S$ be the sphere centered at the origin with radius $12$. What is the triple integral of the scalar field $f(x, y, z) = z - x^2$ over $S$ in spherical coordinates? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^\pi \int_0^{2\pi} \int_{-6}^{6} \rho^3\cos(\varphi)\sin(\varphi) - \rho^4\cos^2(\theta)\sin^3(\varphi) \cdots \, d\rho \, d\theta \, d\varphi$ (Choice B) B $\dfrac{1}{2} \int_0^\pi \int_0^{2\pi} \int_{-6}^{6} \rho^3\sin(2\varphi) - \rho^4\cos(\theta)\sin^3(\varphi) \cdots \, d\rho \, d\theta \, d\varphi$ (Choice C) C $ \int_0^\pi \int_0^{2\pi} \int_0^{12} \rho^3\cos(\varphi)\sin(\varphi) - \rho^4\cos^2(\theta)\sin^3(\varphi) \cdots \, d\rho \, d\theta \, d\varphi$ (Choice D) D $ \int_0^\pi \int_0^{2\pi} \int_0^{12} \rho\cos(\varphi) - \rho^2\cos^2(\theta)\sin^2(\varphi) \cdots \, d\rho \, d\theta \, d\varphi$
Answer: The bounds we'll use are $0 < \theta < 2\pi$ and $0 < \varphi < \pi$. Here is the change of variables for spherical coordinates. $\begin{aligned} x &= \rho \cos(\theta) \sin(\varphi) \\ \\ y &= \rho \sin(\theta) \sin(\varphi) \\ \\ z &= \rho \cos(\varphi) \end{aligned}$ We want to represent the sphere $S$ with bounds in spherical coordinates. The standard unit sphere needs $\varphi$ to range across $[0, \pi]$, $\theta$ to range across $[0, 2\pi]$, and $\rho$ to range across $[0, 1]$. The sphere $S$ has radius $12$, so it needs $\rho$ to range across $[0, 12]$ : $ \int_0^\pi \int_0^{2\pi} \int_0^{12} \cdots \, d\rho \, d\theta \, d\varphi$ We can now put $f(x, y, z)$ in the integrand, but we need to substitute $x$, $y$, and $z$ for their definitions in spherical coordinates. $ \int_0^\pi \int_0^{2\pi} \int_0^{12} \rho\cos(\varphi) - \rho^2\cos^2(\theta)\sin^2(\varphi) \cdots \, d\rho \, d\theta \, d\varphi$ The final step is finding the Jacobian of spherical coordinates, which we'll need to multiply in to get the final integral. $J(\rho, \theta, \varphi) = \rho^2\sin(\varphi)$ [Derivation] The integral in spherical coordinates: $ \int_0^\pi \int_0^{2\pi} \int_0^{12} \rho^3\cos(\varphi)\sin(\varphi) - \rho^4\cos^2(\theta)\sin^3(\varphi) \cdots \, d\rho \, d\theta \, d\varphi$